Báo cáo toán học: "The polynomial part of a restricted partition function related to the Frobenius problem" pptx

The polynomial part of a restricted partition functionrelated to the Frobenius problem Matthias Beck Department of Mathematical Sciences State University of New York Binghamton, NY 13902

The polynomial part of a restricted partition function

related to the Frobenius problem Matthias Beck

Department of Mathematical Sciences

State University of New York

Binghamton, NY 13902–6000, USA

matthias@math.binghamton.edu

Ira M Gessel ∗ Department of Mathematics Brandeis University Waltham, MA 02454–9110, USA gessel@brandeis.edu

Takao Komatsu Faculty of Education Mie University Mie, 514–8507, Japan komatsu@edu.mie-u.ac.jp

Submitted: May 29, 2001; Accepted: September 4, 2001

MR Subject Classifications: Primary 05A15; Secondary 11P81, 05A17

Abstract

Given a set of positive integers A = {a1, , a n }, we study the number p A(t) of

nonnegative integer solutions (m1, , m n) toPn

j=1 m j a j =t We derive an explicit

formula for the polynomial part of p A

Let A = {a1, , a n } be a set of positive integers with gcd(a1, , a n) = 1 The

classical Frobenius problem asks for the largest integer t (the Frobenius number ) such

that

m1a1+· · · + m n a n = t has no solution in nonnegative integers m1, , m n For n = 2, the Frobenius number is (a1 − 1)(a2− 1) − 1, as is well known, but the problem is extremely difficult for n > 2.

(For surveys of the Frobenius problem, see [R, Se].) One approach [BDR, I, K, S ¨O] is to

study the restricted partition function p A (t), the number of nonnegative integer solutions (m1, , m n) toPn

j=1 m j a j = t, where t is a nonnegative integer The Frobenius number

is the largest integral zero of p A (t) Note that, in contrast to the Frobenius problem, in the definition of p A we do not require a1, , a n to be relatively prime In the following,

a1, , a n are arbitrary positive integers.

∗Research partially supported by NSF grant DMS-9972648.

It is clear that p A (t) is the coefficient of z t in the generating function

(1− z a1)· · · (1 − z a n).

If we expand G(z) by partial fractions, we see that p A (t) can be written in the form

X

λ

P A,λ (t)λ t ,

where the sum is over all complex numbers λ such that λ a i = 1 for some i, and P A,λ (t) is

a polynomial in t The aim of this paper is to give an explicit formula for P A,1 (t), which

we denote by P A (t) and call the polynomial part of p A (t) It is easy to see that P A (t) is

a polynomial of degree n − 1 (More generally, the degree of P A,λ (t) is one less than the number of values of i for which λ a i = 1.) It is well known [PS, Problem 27] that

n−1

(n − 1)! a1· · · a n

+ O t n−2

.

Our theorem is a refinement of this statement We note that Israilov derived a more

complicated formula for P A (t) in [I].

Let us define Q A (t) by p A (t) = P A (t) + Q A (t) From the partial fraction expansion above, it is clear that Q A (and hence also p A ) is a quasi-polynomial, that is, an expression

of the form

c d (t)t d+· · · + c1(t)t + c0(t),

where c0, , c d are periodic functions in t (See, for example, Stanley [St, Section 4.4], for more information about quasi-polynomials.) In the special case in which the a i are

pairwise relatively prime, each P A,λ (t) for λ 6= 1 is a constant, and thus Q A (t) is a periodic function with average value 0, and this property determines Q A (t), and thus

P A (t) Discussions of Q A (t) can be found, for example, in [BDR, I, K].

We define the Bernoulli numbers B j by

z

e z − 1 =

X

j≥0

B j z

j

(so B0 = 1, B1 =−1

2, B2 = 16, B4 =−1

30, and B n = 0 if n is odd and greater than 1.)

Theorem.

P A (t) = 1

a1· · · a n

n−1

X

m=0

(−1) m

(n − 1 − m)!

X

k1+···+k n =m

a k1

1 · · · a k n

n

B k1· · · B k n

k1!· · · k n! t

n−1−m (2)

a1· · · a n

n−1

X

m=0

(−1) m

(n − 1 − m)!

k1+2k2···+mk m =m

(−1) k2+···+k m

k1!· · · k m!



B1s1

1· 1!

k1

· · ·



B m s m

m · m!

k m

t n−1−m , (3)

where s i = a i1+· · · + a i

n

As an intermediate step, we first prove the following more elegant but less explicit

formula for P A (t).

Proposition P A (t) is the constant term in z in

(1− e a1z)· · · (1 − e a n z) .

Proof As noted earlier, p A (t) is the coefficient of z t in the generating function

(1− z a1)· · · (1 − z a n).

Hence if we let f (z) = G(z)/z t+1 then p A (t) = Res(f (z), z = 0) As in [BDR], we use the residue theorem to derive a formula for p A (t) Since clearly lim

R→∞

Z

|z|=R

f (z) dz = 0,

p A (t) = − Res(f(z), z = 1) −XRes(f (z), z = λ).

Here the sum is over all nontrivial a1, , a n th roots of unity λ It is not hard to see that Res(f (z), z = λ) may be expressed in the form u λ (t)λ −t for some polynomial u λ (t),

and thus it follows from our earlier discussion that− Res(f(z), z = λ) = P A,λ −1 (t)λ −t In particular,

P A (t) = − Res(f(z), z = 1).

To compute this residue, note that

Res(f (z), z = 1) = Res(e z f (e z ), z = 0),

so that

P A (t) = − Res



e −tz

(1− e a1z)· · · (1 − e a n z), z = 0



2 Proof of the theorem The coefficient of t n−1−m in P A (t) is by (4) the coefficient of z −n+m

in

(−1) n−m

(n − 1 − m)! ·

1 (1− e a1z)· · · (1 − e a n z),

which is the coefficient of z m in

(−1) m

(n − 1 − m)! a1· · · a n

where B(z) = z/(e z − 1), and this implies (2).

To prove (3), we first note that

log



z

e z − 1



j≥1

(−1) j−1 B j

j

z j

j! ,

as can easily be proved by differentiating both sides Then

B(a1z) · · · B(a n z) = expX

j≥1

(−1) j−1 B j s j

j

z j

j!

j≥1

exp

 (−1) j−1 B j s j

j

z j

j!



Since B 2i+1 = 0 for i > 0, ( −1) j−1 B j =−B j for j > 1, and (3) follows from (5) and (6).

2 Remark It is possible to avoid the use of complex analysis and give a purely formal power

series proof of the theorem We indicate here how this can be done We work with formal Laurent series, which are power series with finitely many negative powers of the variable

If F (z) =P

i u i z i is a formal Laurent series (u i is nonzero for only finitely many negative

values of i) then the residue of F (z) is res F (z) = u −1 An elementary fact about formal

Laurent series is the change of variables formula for residues: If g(z) is a formal power series with g(0) = 0 and g 0(0)6= 0 then

res F (z) = res F (g(z))g 0 (z).

(See, for example, Goulden and Jackson [GJ, p 15].)

By partial fractions, we have

(1− z a

1)· · · (1 − z a m) =

c1

1− z +· · · +

c m

(1− z) m + R(z), where R(z) is a rational function of z with denominator not divisible by 1 − z It follows

from our earlier discussion that

∞

X

t=0

P A (t)z t = c1

1− z +· · · +

c m

(1− z) m

and thus

P A (t) =

∞

X

l=1

c l



t + l − 1

l − 1



,

where we take c l to be 0 for l > m Now let U (z) = G(1 − z) Then

(1− (1 − z) a1)· · · (1 − (1 − z) a m)

= c1

z +· · · + c m

z m + R(1 − z),

where R(1 −z) has a formal power series expansion (with no negative powers of z), and thus

c l = res z l−1 U (z) Note that this holds for all l ≥ 1, since for l > m, c l = res z l−1 U (z) = 0.

P A (t) =

∞

X

l=1

c l



t + l − 1

l − 1



= resU (z)

z

m

X

l=1

z l



t + l − 1

l − 1



= res U (z) (1− z) t+1

We now apply the change of variables formula with g(z) = 1 − e z and we obtain

P A (t) = − res U (1 − e z)

e tz

(1− e a1z)· · · (1 − e a m z), which is (4), and the proof continues as before

References

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